CodeAlong: From Distributions to Hypotheses

Learning Objectives

• To be able to use probability density functions to calculate probability of specific values.

• To identify normally distributed features.

• To perform a hypothesis test to compare numeric data between 2 groups.

import pandas as pd
import numpy as np

import matplotlib as mpl
import matplotlib.pyplot as plt
import seaborn as sns
from scipy import stats

sns.set_context('talk')
mpl.rcParams['figure.figsize'] = [12,6]

Exploring Distributions

Dataset: https://archive.ics.uci.edu/ml/datasets/student+performance

pd.set_option('display.max_columns',100)

## read in the Data/student/student-mat.csv (it uses ";" as the sep)

# display info and .head

## Calculate an Avg Grade column by averaging G1, G2,G3, 
# then divide by 20, and * 100 (to make %'s')

## plot the distribution of Avg Grade 

Is it normally distributed?

## use scipy's normaltest

• We have our p-value for our normaltest, but what does it mean??
  • Check the docstring for the normaltest to find out the null hypothesis of the test.

Calculating Probabilities with Scipy's Probability Density Functions

## Get the mean, std, min, and max for the Avg Grade column

## generate a linearly-spaced array of values that span the min to the max

## use stats.norm.pdf to get the PDF curve that corresponds to your distribution's values

## Plot the histogram again AND then plot the pdf we calculated.

Looks pretty normal! But can we confirm for a fact that its normal?

Q1: what is the probability of a student getting a score of 90 or above?

## Plot the histogram again AND pdf again

## Add a vpsan to the plot showing the region we want to calc prob for

How can we calculate this probability? Can we use the PDF?

## try making a list of values from 90-100 and getting the pdf values


## Sum the values to get the total probability. 

Whats the flaw to this approach?

## Use the cumulative density function to find prob of 90 OR lower.

Now, we want the opposite probability, probability of being GREATER Than 90.

# calc 1-prob of 90 or lower.

• Answer: there is a 2.4% chance of having a score greater than 90.

Hypothesis Testing

Q: Do students with internet access have different average grades than students who do not have internet access?

State The Hypothesis

• $H_0$ (Null Hypothesis): Students with internet access have the same average grades as students who do not.
• $H_A$ (Alternative Hypothesis): Students with internet access have significantly different average grades compared to students who do not.

Visualize and Separate Groups

• Visualize the histogram of Avg Grade again, but separate it into groups based on the "internet" column.
• Note: when comparing 2 groups with seaborn's histplot, you will want to add common_norm=False

## visualize the histobram of Avg Grade again, but separate it by "internet"

## Plot a bar plot of the Avg Grade for students with internet vs those that do not have it

## Separate the 2 groups into 2 varaibles

T-Test Assumptions

• Since we are comparing a numeric measurement between 2 groups, we want to run a 2-sample (AKA independent T-test).

• The Assumptions are:

  • No significant outliers
  • Normality
  • Equal Variance

Assumption: No Sig. Outliers

## check yes group for outliers using z-score >3 rule.

## check no group for outliers using z-score >3 rule.

No outliers to worry about! Assumption met.

Assumption: Normally Distributed Groups

## use normaltest to check if yes group is normally distributed

## use normaltest to check if no group is normally distributed

• Did we meet the assumption of normality?

Assumption: Equal Variance

## use Levene's test to check if groups have equal variance

Did we meet the assumption of equal variance?

Perform Final Hypothesis Test (T-Test)

• Since we met all of the assumptions for the test we can proceed with our t-test.
  • Next class we will discuss what we would do if we did NOT meet the assumptions.

## run stats.ttest_ind on the 2 groups

What is our p-value? Is it less than our alpha of .05? What does this mean?

Our T-Test returned a p-value of ____. Since p </>.05, we can reject/fail to reject the null hypothesis that students with internet access have the same average grades as students who do not.

We therefore conclude that there is/is not a significant difference in Average Grades between students who do/do not have internet access.

Our visualization below shows that students with internet access have HIGHER/LOWER/EQUAL average grades.

## Add a summary visual to support our results.